= Solution
The unrestricted <maximum-likelihood estimators> remain $(\bar X,\bar Y)$. Subject to the null constraint, minimizing $(\mu_X-\bar X)^2$ on $[A,\infty)$ gives $\widehat\mu_X=\max(A,\bar X)$, while $\widehat\mu_Y=0$. Consequently
$$
\boxed{T=-2\log\Lambda=n[\bar Y^2+(A-\bar X)_+^2],\qquad \text{reject if }T>c_\alpha.}
$$
We must calibrate this <likelihood-ratio test> over the entire composite null. Under that null let $W=\sqrt n(\bar X-A)$ and $Z=\sqrt n\bar Y$. Then $W\sim N(\delta,1)$ with $\delta=\sqrt n(\mu_X-A)\geq0$, while $Z\sim N(0,1)$ independently, and $T=Z^2+(-W)_+^2$. Couple all null distributions by $W=W_0+\delta$, with a fixed standard Gaussian $W_0$. Increasing $\delta$ decreases $(-W_0-\delta)_+^2$ pointwise, so the largest rejection probability occurs at $\delta=0$, namely $\mu_X=A$.
At this boundary the sign of $W$ is independent of $W^2$, with each sign having probability one half. On $W\geq0$, $T=Z^2\sim\chi_1^2$; on $W<0$, $T=Z^2+W^2\sim\chi_2^2$. Thus the least favourable <null hypothesis> law is the exact <chi-bar-squared distribution> $\tfrac12\chi_1^2+\tfrac12\chi_2^2$. Choose $c_\alpha>0$ by
$$
\boxed{\frac12\mathbb P(\chi_1^2>c_\alpha)+\frac12\mathbb P(\chi_2^2>c_\alpha)=\alpha,\quad\text{equivalently }1-\Phi(\sqrt{c_\alpha})+\frac12e^{-c_\alpha/2}=\alpha.}
$$
The tail is a <continuous function> and strictly decreasing from one to zero, so this threshold is unique for $0<\alpha<1$. The null rejection probability is at most $\alpha$ and equals $\alpha$ at $\mu_X=A$. This proves the required size, rather than using an interior chi-squared approximation.
To compare, put $c_1=z_{1-\alpha/2}^2$, the threshold in (i). Since $\chi_2^2$ is stochastically larger than $\chi_1^2$ (add an independent squared standard Gaussian), the mixture threshold obeys $c_\alpha>c_1$. When $\bar X\geq A$, test (ii) therefore requires stronger evidence from $\bar Y$ than test (i). When $\bar X<A$, test (ii) also responds to violation of the mean constraint and can reject even with $\bar Y=0$. \b[Neither critical region contains the other.] In the coordinates $(W,Z)$, the acceptance region in (i) is the strip $|Z|\leq\sqrt{c_1}$; in (ii) it is the right half-strip $W\geq0$, $|Z|\leq\sqrt{c_\alpha}$ together with the left half-disk $W<0$, $W^2+Z^2\leq c_\alpha$. The extra constraint changes both the statistic and its correct null calibration.
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